$A$ charge $Q$ is placed at the centre of a cube of side $a$. The total flux of electric field through the six surfaces of the cube is

  • A
    $\frac{6 Q a^2}{\epsilon_0}$
  • B
    $\frac{Q a^2}{6 \epsilon_0}$
  • C
    $Q / \epsilon_0$
  • D
    $Q a^2 / \epsilon_0$

Explore More

Similar Questions

If a charge $Q$ is placed at the corner of a cube,what is the electric flux passing through one of its faces?

In the figure,a $+Q$ charge is located at one of the corners of the cube. What is the electric flux through the cube due to the $+Q$ charge?

Discuss some points about Gauss's law.

Difficult
View Solution

Let the electrostatic field $E$ at distance $r$ from a point charge $q$ not be an inverse square but instead an inverse cubic, e.g., $E = k \cdot \frac{q}{r^3} \hat{r}$, where $k$ is a constant.
Consider the following two statements:
$(I)$ Flux through a spherical surface enclosing the charge is $\phi = q_{\text{enclosed}} / \varepsilon_0$.
$(II)$ $A$ charge placed inside a uniformly charged shell will experience a force.
Which of the above statements are valid?

Two surfaces $A$ and $B$ are enclosing the charges as shown below. The total normal electric induction ($T$.$N$.$E$.$I$) through the surfaces $A$ and $B$ are respectively.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo